Hi there!
Recall that:
[tex]\Phi_E = \oint E \cdot dA = \frac{Q_{enclosed}}{\epsilon_{0}}[/tex]
The electric flux is PROPORTIONAL to the total ENCLOSED charge and inversely proportional to the permittivity of free space.
Using the equivalent equation:
[tex]\Phi_E = \frac{Q_{enclosed}}{\epsilon_0}[/tex]
ε₀ is a constant, so the electric flux is only dependent on the total enclosed charge. Increasing the enclosed charge increases the electric flux and vice-versa.
